NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1

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H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1

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H3 – Access NCERT Solutions for Class 11 Maths Chapter 15 –Statistics

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H3 – Exercise 15.1

There are a total of twelve questions in Class 11 Maths Ch 15 Ex 15.1. These problems are based on the Measures of dispersion. Sometimes students come across difficulties during solving the questions. To understand the term with examples, they must look up the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1. It contains simplified formulas or detailed theories. So, learners can grasp properly.

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H3 – NCERT Solutions for Class 11 Maths Chapters

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The following are the NCERT Solutions for Class 11 Maths Chapters:There are 16 chapters in total.

Chapter 1 – Sets

Sets and their representation are covered in this chapter. The empty set, Finite and Infinite sets, Equal sets, Subsets, Power sets, and Universal sets are the topics covered in this chapter. The chapter also teaches students how to create Venn diagrams and introduces them to the ideas of set intersection and union.

Chapter 2 – Relations and Functions

In this chapter, the terms ordered pairs, a cartesian product of sets, number of elements in the cartesian product of two finite sets, the definition of relation, graphical diagrams, domain, co-domain, and range of a relation are taught to students.

Chapter 3 -Trigonometric Functions

The chapter covers positive and negative angles, how to measure angles in radians and degrees, and how to convert between different measures. The definition of trigonometric functions using the unit circle, the general solution of trigonometric equations, the signs, the domain, the range, and their graphs are all covered in this chapter.

Chapter 4 – Principle of Mathematical Induction

The chapter on the “Principle of Mathematical Induction,” which covers a variety of issues, looks at natural numbers as the least inductive subset of real numbers to demonstrate the induction and provide motivation for its use.

Chapter 5 – Complex Numbers and Quadratic Equations

The chapter goes into great length regarding why it is necessary for complex numbers, especially √−1, to be justified by the difficulty in resolving some quadratic equations. Additionally, the chapter allows the students to study the polar representation of complex numbers, the argand plane, and the algebraic features of complex numbers.

Chapter 6- Linear Inequalities

As the title indicates, the chapter Linear Inequalities discusses the concept of a linear inequality. This chapter also discusses the concepts of algebraic solutions of linear inequalities in one variable and their representation on the number line, graphical representations of linear inequalities in two variables, as well as the graphical method of solving a system of linear inequalities in two variables.

Chapter 7- Permutations and Combinations

The chapter covers the fundamental terms of counting, factorial n. (n! ), permutations, combinations, formula derivation, linkages between formulae, and straightforward applications.

Chapter 8- Binomial Theorem

Students get to know about the Binomial Theorem for positive integers. Complex computations that are challenging to answer with repeated multiplication can be solved using this theorem. The history, formulation, and demonstration of the Binomial Theorem for positive integral indices are covered in this chapter.

Chapter 9- Sequences and Series

Sequence and series, an arithmetic progression (A. P.), arithmetic mean (A. M.), geometric progression (G. P.), general term of a G. P., the sum of first n terms of a G. P., infinite G. P., and its sum, geometric mean (G. M.), the relationship between A. M. and G. M, formulas for the sums of special series, and other topics are covered in this chapter.

Chapter 10- Straight Lines

The chapter aids learners in remembering two-dimensional geometry from prior courses. The chapter includes discussions on shifting the origin, the slope of a line, and the angle between two lines, several equational forms for lines, including those parallel to axes, slope-intercept form, point-slope form, intercept form, two-point form and normal form.

Chapter 11- Conic Sections

Conic Sections as a topic are thoroughly covered in this chapter. A conic section in Mathematics is a curve formed by the surface of a cone and a plane intersecting. The subjects discussed include the degenerate cases of conic sections, such as the circle, ellipse, parabola, hyperbola, point, straight line, and two intersecting lines.

Chapter 12- Introduction to Three-Dimensional Geometry

Each geographical point has three coordinates. Students in Class 11 will master the fundamentals of three-dimensional geometry in this chapter. The theory of coordinate axes and planes in three dimensions, as well as the coordinates of a point, the distance between two points, and the section formula, are all thoroughly covered in this chapter.

Chapter 13- Limits and Derivatives

The Limits and Derivatives chapter offers a basic overview of calculus. Calculus is a field of Mathematics that studies how the points in a domain change and how the value of a function changes. Students initially get an intuitive understanding of derivatives in this chapter.

Chapter 14- Mathematical Reasoning

The basics of deductive reasoning are discussed in this chapter. Connecting words and phrases, mathematically valid propositions, and other subjects are covered in this chapter.

Chapter 15- Statistics

Statistics, as students are aware, deals with information gathered for specific purposes.. Learners will study some of the key measures of dispersion in this chapter, along with how to calculate them for both ungrouped and grouped data. Measures of dispersion, range, mean deviation, variance, standard deviation, and analysis of frequency distributions with similar means, but distinct variances are among the subjects covered in this chapter by the students. By inculcating the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 along with exercises. It will provide a conceptual understanding orsolidify the foundational part. . By completing the three activities and another activity in the chapter, students can have a deeper understanding of the material covered in the chapter. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 is available to download from the Extramarks website or mobile app. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 will also be beneficial for understanding advanced statistical concepts.

Chapter 16- Probability

The term “probability” is discussed throughout the chapter as a way to gauge how uncertain particular phenomena are, or just the likelihood that an event will occur.

H3 – NCERT Solution Class 11 Maths of Chapter 15 All Exercises

Students can access the NCERT Solutions Class 11 Maths of Chapter 15 All Exercises can be accessed by students on the Extramarks platform.

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

There are a total of twelve problems in the NCERT textbook of Class 11. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 help students in understanding topics and working through challenging issues at their speed. The NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1 assists students in identifying gaps between course information and written tasks.

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H3 – Important Topics Covered in Exercise 15.1 of Class 11 Maths NCERT Solutions

The mean deviation, a measure of dispersion, is the main topic of the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1. The four basic categories of dispersion measures are range, standard deviation, and quartile deviation. Any given data range offers us a general understanding of variability, but it does not reveal how widely the data are dispersed from a measure of central tendency. Mathematicians employ the mean and standard deviation as a result. Both grouped and ungrouped data can be used for these measurements. Combined data may be divided into two categories: continuous frequency distribution and discrete frequency distribution. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 also helps students to save precious time.

Including the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 while solving the Ch 15 Maths Class 11 Ex 15.1. It will build a strong foundation for the concepts of Statistics.

Finding the mean deviation from the mean and the mean deviation from the median is the focus of this activity. The student’s understanding of the concepts will improve, providing a solid foundation for understanding advanced statistics topics.Extramarks provides NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1.

There are many benefits to using the NCERT solution. The solutions provided by Extramarks are categorised by chapter exercises, such as the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1, so students are not confused. NCERT Textbooks and NCERT Solutions are essential resources for the CBSE board examinations. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 and other exercise solutions make learning easier for Class 11 students who are already dealing with a lot of hurdles. Here, the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 are created with the utmost care and focus by Extramarks’ special faculty. Learn from the best teachers and experienced minds. Use the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 and other exercises to review and clear their doubts.

Q.1 Find the mean deviation about the mean for the data in Exercises 1 and 2.
Q.1: 4, 7, 8, 9, 10, 12, 13, 17
Q.2: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Ans.
1.

$\begin{array}{l} Mean = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8} \\ = \frac{80}{8} \\ = 10 \\ Mean deviation = {\frac{\begin{array}{l} | 4 - 10 | + | 7 - 10 | + | 8 - 10 | + | 9 - 10 | + | 10 - 10 | \\ + | 12 - 10 | + | 13 - 10 | + | 17 - 10 | \end{array}}{8}} \\ = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8} \\ = \frac{24}{8} \\ = 3 \end{array}$

Thus, the mean deviation is 3. 2.

$\begin{array}{l} Mean = \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10} \\ = \frac{500}{10} \\ = 50 \\ Mean deviation = {\frac{\begin{array}{l} | 38 - 50 | + | 70 - 50 | + | 48 - 50 | + | 40 - 50 | + \\ | 42 - 50 | + | 55 - 50 | + | 63 - 50 | + | 46 - 50 | + \\ | 54 - 50 | + | 44 - 50 | \end{array}}{10}} \\ = \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10} \\ = \frac{84}{10} = 8 .4 \\ Thus, the mean deviation is 8.4 . \end{array}$

Q.2 Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Ans.

$\begin{array}{l} 3. A scending order of data: \\ 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 \\ Here, n = 12 (even) \\ Median = \frac{{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term}{2} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{{(\frac{12}{2})}^{th} term + {(\frac{12}{2} + 1)}^{th} term}{2} \\ = \frac{6^{th} term + 7^{th} term}{2} \\ = \frac{13 + 14}{2} \\ = \frac{27}{2} \\ Median = 13 .5 \\ Mean deviation about median \\ = {\frac{\begin{array}{l} | 10 - 13 .5 | + | 11 - 13 .5 | + | 11 - 13 .5 | + | 12 - 13 .5 | + \\ | 13 - 13 .5 | + | 13 - 13 .5 | + | 14 - 13 .5 | + | 16 - 13 .5 | + \\ | 16 - 13 .5 | + | 17 - 13 .5 | + | 17 - 13 .5 | + | 18 - 13 .5 | \end{array}}{12}} \\ = \frac{(\begin{array}{l} 3 .5 + 2 .5 + 2 .5 + 1 .5 + 0 .5 + 0 .5 + 0 .5 + 2 .5 + 2 .5 \\ + 3 .5 + 3 .5 + 4 .5 \end{array})}{12} \\ = \frac{28}{12} \\ = 2 .33 (Approx .) \end{array}$ $\begin{array}{l} 4 . Ascending order of data: \\ 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} Here, n = 10 (even) \\ Median = \frac{{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term}{2} \\ = \frac{{(\frac{10}{2})}^{th} term + {(\frac{10}{2} + 1)}^{th} term}{2} \\ = \frac{5^{th} term + 6^{th} term}{2} \\ = \frac{46 + 49}{2} \\ = \frac{95}{2} \\ Median = 47 .5 \\ Mean deviation about median \\ = {\frac{\begin{array}{l} | 36 - 47 .5 | + | 42 - 47 .5 | + | 45 - 47 .5 | + | 46 - 47 .5 | + \\ | 46 - 47 .5 | + | 49 - 47 .5 | + | 51 - 47 .5 | + | 53 - 47 .5 | + \\ | 60 - 47 .5 | + | 72 - 47 .5 | \end{array}}{10}} \\ = \frac{(\begin{array}{l} 11 .5 + 5 .5 + 2 .5 + 1 .5 + 1 .5 + 1 .5 + 3 .5 + 5 .5 \\ + 12 .5 + 24 .5 \end{array})}{10} \\ = \frac{70}{10} = 7 \end{array}$

Q.3 5. Find the mean deviation about the mean for the data in Exercises 5 and 6.

Xi	5	10	15	20	25
fi	7	4	6	3	5

Xi	10	30	50	70	90
fi	4	24	28	16	8

Ans.

$\begin{matrix} x_{i} & f_{i} & f_{i} x_{i} & |x_{i} - \bar{x}| & f_{i} |x_{i} - \bar{x}| \\ 5 & 7 & 35 & |5 - 14| = 9 & 63 \\ 10 & 4 & 40 & |10 - 14| = 4 & 16 \\ 15 & 6 & 90 & |15 - 14| = 1 & 6 \\ 20 & 3 & 60 & |20 - 14| = 6 & 18 \\ 25 & 5 & 125 & |25 - 14| = 11 & 55 \\ Total & 25 & 350 & 158 \end{matrix}$

$\begin{array}{l} Mean (\bar{x}) = \sum f_{i} x_{i} \end{array}$

$\begin{array}{l} \sum f_{i} \\ = \frac{350}{25} \\ = 14 \\ Mean deviation about mean \\ = \frac{\sum f_{i} |x_{i} - \bar{x}|}{\sum f_{i}} \\ = \frac{158}{25} = 6 .32 \\ Thus, the mean deviation about mean is 6.32. \end{array}$

$\begin{matrix} x_{i} & f_{i} & f_{i} x_{i} & |x_{i} - \bar{x}| & f_{i} |x_{i} - \bar{x}| \\ 10 & 4 & 40 & |10 - 50| = 40 & 160 \\ 30 & 24 & 720 & |30 - 50| = 20 & 480 \\ 50 & 28 & 1400 & |50 - 50| = 0 & 0 \\ 70 & 16 & 1120 & |70 - 50| = 20 & 320 \\ 90 & 8 & 720 & |90 - 50| = 40 & 320 \\ Total & 80 & 4000 & 1280 \end{matrix}$

$\begin{array}{l} Mean (\bar{x}) = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \\ = \frac{4000}{80} \\ = 50 \\ Mean deviation about mean \\ = \frac{\sum f_{i} |x_{i} - \bar{x}|}{\sum f_{i}} \\ = \frac{1280}{80} = 16 \\ Thus, the mean deviation about mean is 16. \end{array}$

Q.4 Find the mean deviation about the median for the data in Exercises 7 and 8.

Xi	5	7	9	10	12	15
fi	8	6	2	2	2	6

Xi	15	21	27	30	35
fi	3	5	6	7	8

Ans.

\begin{matrix} x_{i} & f_{i} & C . f . & | x_{i} - Median | & f_{i} | x_{i} - Median | \\ 5 & 8 & 8 & | 5 - 7 | = 2 & 16 \\ 7 & 6 & 14 & | 7 - 7 | = 0 & 0 \\ 9 & 2 & 16 & | 9 - 7 | = 2 & 4 \\ 10 & 2 & 18 & | 10 - 7 | = 3 & 6 \\ 12 & 2 & 20 & | 12 - 7 | = 5 & 10 \\ 15 & 6 & 26 & | 15 - 7 | = 8 & 48 \\ Total & 26 & 84 \end{matrix}

$\begin{array}{l} n = 26 (even) \\ Median = \frac{{(\frac{n}{2})}^{th} term+ {(\frac{n}{2} +1)}^{th} term}{2} \\ = \frac{{(\frac{26}{2})}^{th} term+ {(\frac{26}{2} +1)}^{th} term}{2} \\ = \frac{13^{th} {term+14}^{th} term}{2} \\ = \frac{7+7}{2} \\ = 7 \\ Mean deviation about median \\ = \frac{\sum f_{i} |x_{i} -Me|}{\sum f_{i}} \\ = \frac{84}{26} \\ = 3.23 \\ Thus, Mean deviation about median is 3.23. \end{array}$

$\begin{matrix} x_{i} & f_{i} & C . f . & |x_{i} - Median| & f_{i} |x_{i} - Median| \\ 15 & 3 & 3 & |15 - 30| = 15 & 45 \\ 21 & 5 & 8 & |21 - 30| = 9 & 45 \\ 27 & 6 & 14 & |27 - 30| = 3 & 18 \\ 30 & 7 & 21 & |30 - 30| = 0 & 0 \\ 35 & 8 & 29 & |35 - 30| = 5 & 40 \\ Total & 29 & 148 \end{matrix}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} n = 29 (odd) \\ Median= {(\frac{n+1}{2})}^{th} term \\ = {(\frac{29+1}{2})}^{th} term \\ {= 15}^{th} term \\ = 30 \\ Mean deviation about Median \\ = \frac{\sum f_{i} |x_{i} -Me|}{\sum f_{i}} \\ = \frac{148}{29} \\ = 5.1 \\ Thus, the Mean deviation about Median is 5.1. \end{array}$

Q.5 Find the mean deviation about the mean for the data in Exercises 9 and 10.

Income per day	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Number of persons	4	8	9	10	7	5	4	3

10.

Height in cms	95-105	105-115	115-125	125-135	135-145	145-155
Number of boys	9	13	26	30	12	10

Ans.

$\begin{matrix} Income & f_{i} & x_{i} & \begin{array}{l} u_{i} \\ = \frac{x - A}{h} \end{array} & f_{i} u_{i} & |x_{i} - \bar{x}| & f_{i} |x_{i} - \bar{x}| \\ 0 - 100 & 4 & 50 & - 4 & - 16 & |50 - 358| = 308 & 1232 \\ 100 - 200 & 8 & 150 & - 3 & - 24 & |150 - 358| = 208 & 1664 \\ 200 - 300 & 9 & 250 & - 2 & - 18 & |250 - 358| = 108 & 972 \\ 300 - 400 & 10 & 350 & - 1 & - 10 & |350 - 358| = 8 & 80 \\ 400 - 500 & 7 & 450 = A & 0 & 0 & |450 - 358| = 92 & 644 \\ 500 - 600 & 5 & 550 & 1 & 5 & |550 - 358| = 192 & 960 \\ 600 - 700 & 4 & 650 & 2 & 8 & |650 - 358| = 292 & 1168 \\ 700 - 800 & 3 & 750 & 3 & 9 & |750 - 358| = 392 & 1176 \\ Total & 50 & - 46 & 7896 \end{matrix}$

$\begin{array}{l} Mean = A + h \times \frac{\sum f_{i} u_{i}}{\sum f_{i}}, {where u}_{i} = \frac{x_{i} - A}{h} and h = 100 \\ = 450 + 100 \times \frac{- 46}{50} \\ = 450 - 2 \times 46 \\ = 450 - 92 \\ = 358 \\ Mean deviation about mean \\ = \frac{\sum f_{i} |x_{i} - \bar{x}|}{\sum f_{i}} \\ = \frac{7896}{50} \\ = 157 .92 \end{array}$

10.

$\begin{matrix} \begin{array}{l} Class - \\ interval \end{array} & f_{i} & x_{i} & u_{i} = \frac{x_{i} - A}{h} & f_{i} u_{i} & |x_{i} - \bar{x}| & f_{i} |x_{i} - \bar{x}| \\ 95 - 105 & 9 & 100 & - 3 & - 27 & |100 - 125 .3| = 25 .3 & 227 .7 \\ 105 - 115 & 13 & 110 & - 2 & - 26 & |110 - 125 .3| = 15 .3 & 196 .9 \\ 115 - 125 & 26 & 120 & - 1 & - 26 & |120 - 125 .3| = 5 .3 & 137 .8 \\ 125 - 135 & 30 & 130 = A & 0 & 0 & |130 - 125 .3| = 4 .7 & 141 \\ 135 - 145 & 12 & 140 & 1 & 12 & |140 - 125 .3| = 14 .7 & 176 .4 \\ 145 - 155 & 10 & 150 & 2 & 20 & |150 - 125 .3| = 24 .7 & 247 \\ Total & 100 & - 47 & 1128 .8 \end{matrix}$

$\begin{array}{l} Mean = A + h \times \frac{\sum f_{i} u_{i}}{\sum f_{i}}, {where u}_{i} = \frac{x_{i} - A}{h} and h = 10 \\ = 130 + 10 \times \frac{- 47}{100} \\ = 130 - 4 .7 \\ = 125 .3 \\ Mean deviation about Mean \\ = \frac{\sum f_{i} |x_{i} - \bar{x}|}{\sum f_{i}} \\ = \frac{1128 .8}{100} \\ = 11 .288 \\ = 11 .29 (Approx) \\ Thus, Mean deviation about Mean is 11.29. \end{array}$

Q.6 Find the mean deviation about median for the following data :

Marks	0-10	Oct-20	20-30	30-40	40-50	50-60
Number of Girls	6	8	14	16	4	2

Ans.

$\begin{array}{l} \begin{matrix} Marks & f_{i} & x_{i} & c . f . & | x_{i} - Me | & f_{i} | x_{i} - \bar{x} | \\ 0 - 10 & 6 & 5 & 6 & | 5 - 27 .85 | = 22 .85 & 137 .1 \\ 10 - 20 & 8 & 15 & 14 & | 15 - 27 .85 | = 12 .85 & 102 .8 \\ 20 - 30 & 14 & 25 & 28 & | 25 - 27 .85 | = 2 .85 & 39 .9 \\ 30 - 40 & 16 & 35 & 44 & | 35 - 27 .85 | = 7 .15 & 114 .4 \\ 40 - 50 & 4 & 45 & 48 & | 45 - 27 .85 | = 17 .15 & 68 .6 \\ 50 - 60 & 2 & 55 & 50 & | 55 - 27 .85 | = 27 .15 & 54 .3 \\ Total & 50 & 517 .1 \end{matrix} \\ N = 50 (even) \\ \frac{N}{2} = \frac{50}{2} = 25 \\ So, the median class is 20 - 30. \\ Median = l + \frac{(\frac{N}{2}) - C}{f} \times h \\ = 20 + \frac{25 - 14}{14} \times 10 [∵ C = 14, f = 14 and h=10] \\ = 20 + 7 .85 = 27 .85 \\ Mean deviation about Meadian \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{517 .1}{50} \\ = 10 .34 \\ Thus, Mean deviation about Meadian is 10.34. \end{array}$

Q.7 Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age	16-20	21-25	36-30	31-35	36-40	41-45	46-50	51-55
Number	5	6	12	14	26	12	16	9

Ans.
The given table can be arranged as given below:

$\begin{matrix} Marks & f_{i} & x_{i} & c . f . & | x_{i} - Me | & f_{i} | x_{i} - \bar{x} | \\ 15 .5 - 20 .5 & 5 & 18 & 5 & | 18 - 38 | = 20 & 100 \\ 20 .5 - 25 .5 & 6 & 23 & 11 & | 23 - 38 | = 15 & 90 \\ 25 .5 - 30 .5 & 12 & 28 & 23 & | 28 - 38 | = 10 & 120 \\ 30 .5 - 35 .5 & 14 & 33 & 37 & | 33 - 38 | = 5 & 70 \\ 35 .5 - 40 .5 & 26 & 38 & 63 & | 38 - 38 | = 0 & 0 \\ 40 .5 - 45 .5 & 12 & 43 & 75 & | 43 - 38 | = 5 & 60 \\ 45 .5 - 50 .5 & 16 & 48 & 91 & | 48 - 38 | = 10 & 160 \\ 50 .5 - 55 .5 & 9 & 53 & 100 & | 53 - 38 | = 15 & 135 \\ Total & 100 & 735 \end{matrix}$

$\begin{array}{l} N = 100 (even) \\ \frac{N}{2} = \frac{100}{2} = 50 \\ So, the median class is 35 .5 - 40 .5. \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} Median = l + \frac{(\frac{N}{2}) - C}{f} \times h \\ = 35 .5 + \frac{50 - 37}{26} \times 5 [∵ C = 37, f = 26 and h = 5] \\ = 35 .5 + 2 .5 = 38 \\ Mean deviation about Meadian \\ = \frac{\sum f_{i} | x_{i} - Me |}{\sum f_{i}} \\ = \frac{735}{100} \\ = 7 .35 \\ Thus, Mean deviation about Meadian is 7.35. \end{array}$

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FAQs (Frequently Asked Questions)

1. What are the important topics discussed in the Class 11 Mathematics Chapter 15?

The topics discussed in the Class 11 Mathematics Chapter 15 are –

Introduction
Methods of Dispersion
Range
Mean Deviation
Variance and Standard Deviation
Analysis of Frequency Distributions

2. What is Statistics?

Statistics is a branch of applied Mathematics that involves describing, collecting, drawing, and analysing conclusions from quantitative data. The mathematical theory behind statistics relies heavily on differential and integral calculus, linear algebra, and probability theory.

3. Are the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 useful to the students?

The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 offer clear explanations in straightforward language to aid students in performing well on examinations. Finding the mean deviation from the mean and mean deviation from the median is the focus of the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1. This solution will aid the students in comprehending the ideas more properly. The time required for preparation can be decreased by using the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 as a quick and effective review tool.

4. Why the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 is crucial for better understanding?

Statistics is one of the most important topics to deal with. The concept of this is not only required for academic examinations but also for competitive examinations. To strengthen the fundamentals of Statistics, learners need to take reference from the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1. Problems are solved systematically odder with accuracy so students do not face any difficulty.

They can download the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 from the online platform Extramakrs. It is also present on the mobile application.

NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1

H3 – Access NCERT Solutions for Class 11 Maths Chapter 15 –Statistics

H3 – Exercise 15.1

H3 – NCERT Solutions for Class 11 Maths Chapters

H3 – NCERT Solution Class 11 Maths of Chapter 15 All Exercises

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

H3 – Important Topics Covered in Exercise 15.1 of Class 11 Maths NCERT Solutions

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1. What are the important topics discussed in the Class 11 Mathematics Chapter 15?

2. What is Statistics?

3. Are the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 useful to the students?

4. Why the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 is crucial for better understanding?

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1

H3 – Access NCERT Solutions for Class 11 Maths Chapter 15 –Statistics

H3 – Exercise 15.1

H3 – NCERT Solutions for Class 11 Maths Chapters

H3 – NCERT Solution Class 11 Maths of Chapter 15 All Exercises

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

H3 – Important Topics Covered in Exercise 15.1 of Class 11 Maths NCERT Solutions

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Please register to view this section

FAQs (Frequently Asked Questions)

1. What are the important topics discussed in the Class 11 Mathematics Chapter 15?

2. What is Statistics?

3. Are the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 useful to the students?

4. Why the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.1 is crucial for better understanding?

Share

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions